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Question

Let N be the smallest positive integer such that N+2N+3N++9N is a number all whose digits are equal. What is the sum of the digits of N?

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Solution

N is an integer.
P=N+2N+3N++9N
P=(1+2+...+9)N=45N
If N is an even integer, then P=45N will end in 0.
This is not possible as all the digits are same.
If N is an odd integer, then P=45N will end in 5.
P=45N=5555a times
Also as P=45N is a multiple of 9.
5 must appear atleast 9 times.
N=55555555545=12345679
Sum of digits of N=37.

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